# Solve $\sin(12º)\sin(24º)\sin(84º-xº) = \sin(30º)\sin(30º)\sin(xº)$

I'm trying to solve this trigonometric equation: $$\sin(12º)\sin(24º)\sin(84º-xº) = \sin(30º)\sin(30º)\sin(xº)$$

I got here after applying Trigonometric Ceva Theorem. Here, I don't know how to solve it, I tried to use $$\sin(xº)\sin(60º-xº)\sin(60º+xº)=\frac{\sin(3xº)}{4}$$, but I couldn't proceed further.

Any hints are appreciated.

• Do you have two $\sin 30^\circ$? Isn't it $\sin^230^\circ=1/4$? – Quang Hoang Dec 9 '19 at 17:35
• Yes, but I put the complete equation after applying the Ceva theorem to give a little bit of context. – Rodrigo Pizarro Dec 9 '19 at 17:36
• With such setups (Trig form of Ceva), you usually have to guess what $x$ is first, and then prove the trigonometric identity. It is often hard to stare and it and see something magically appear, esp working with sin (84-x). (We know a unique solution exists because of monotonicity of sin ( 84 - x) / sin x.) In this case, $x = 18^ \circ$. – Calvin Lin Dec 9 '19 at 17:39
• I knew the answer, I just want to know how to get there after guessing it. – Rodrigo Pizarro Dec 9 '19 at 17:42
• In which case, update your question accordingly. Also, typically you just brute force your way through. The hard part is guessing the value. – Calvin Lin Dec 9 '19 at 17:51

$$\dfrac{\sin x}{\sin(84-x)}=4\sin12\sin24$$
Using the Werner Formulas, $$4\sin y\sin2y=2(\cos y-\cos3y)$$
$$2\cos12-2\cos36=2\cos12-(1+2\cos72)=2\sin42-1$$
$$2\sin42=\dfrac{2\sin42\cos(42-x)}{\sin(84-x)}$$
HINT: Once you get into the form $$\frac{\sin(x)}{\sin(84 - x)}$$ Use the trig identity $$\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B)$$ and then try to cancel the $$\sin(x)$$ in the numerator. I was able to simplify after this, although its a bit of a pain.